Question

${\int }_0^{\pi }\frac{xdx}{{(a^2{\cos }^{2}x+b^2{\sin }^{2}x)}^2}=\frac{{\pi }^2}{k}\frac{(a^2+b^2)}{a^3{b}^3}$. Find the value of $k$.

Answer 1

Let $I={\int }_0^{\pi }\frac{xdx}{{(a^2{\cos }^{2}x+b^2{\sin }^{2}x)}^2}$
Applying integration property $f(a-x)=f\left(x\right)$
$2I={\int }_0^{\pi }\frac{\pi dx}{{(a^2{\cos }^2+x+b^2{\sin }^{2}x)}^2}=2\pi {\int }_0^{\pi /2}\frac{dx}{{(a^2{\cos }^{2}x+b^2{\sin }^{2}x)}^2}$
Using integration property $f(2a-x)=f\left(x\right)$
Therefore
$I=\pi {\int }_0^{\pi /2}\frac{{\sec }^{2}x{\sec }^{2}xdx}{{(a^2+b^2{\tan }^{2}x)}^2}$
Put $b\tan x=a\tan \theta \Rightarrow b{\sec }^{2}xdx=a{\sec }^2\theta d\theta$
$\Rightarrow {\sec }^{2}xdx=\frac{a}{b}(1+{\tan }^2\theta )d\theta$
Therefore
$I=\pi {\int }_0^{\pi /2}\frac{(1+{\tan }^{2}x)\frac{a}{b}(1+{\tan }^2\theta )d\theta }{a^4{(1+{\tan }^2\theta )}^2}$
$=\frac{\pi }{a^{3}b}{\int }_0^{\pi /2}(1+\frac{a^2}{b^2}{\tan }^2\theta ){\cos }^2\theta d\theta$
$=\frac{\pi }{a^3{b}^3}{\int }_0^{\pi /2}(b^2{\cos }^2\theta +a^2{\sin }^2\theta )d\theta$
$=\frac{\pi }{a^3{b}^3}[\frac{1}{2},\frac{\pi }{2}](b^2+a^2)=\frac{{\pi }^2}{4}\frac{a^2+b^2}{a^3{b}^3}$